(→Signal from Question 1) |
(→Signal from Question 1) |
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== Part B == | == Part B == | ||
| − | |||
== Signal from Question 1 == | == Signal from Question 1 == | ||
| − | <math>x(t)=cos(\frac{\pi}{2}t)u(t)\!</math> | + | <math>x(t)=cos(\frac{\pi}{2}t)u(t) = \frac{1}{2}(e^{j0.5\pi t}+e^{-j0.5\pi t})u(t)\!</math> |
| + | |||
| + | <math>y(t) = (3e^{-j0.5\pi} + e^{j1.5\pi} - 1)\times\frac{1}{2}[e^{j0.5\pi t}]u(t)+(3e^{-j0.5\pi} + e^{j1.5\pi} - 1)\times\frac{1}{2}[e^{-j0.5\pi t}]u(t)\!</math> | ||
| + | |||
| + | <math>y(t)=\frac{1}{2}[3e^{j0.5\pi(t-1)}+e^{j0.5\pi(t+3)}-e^{j0.5\pi t}+3e^{-j0.5\pi(t+1)}+e^{j0.5\pi(3-t)}-e^{-j0.5\pi t}]u(t)\!</math> | ||
Latest revision as of 17:37, 26 September 2008
CT LTI signal:
$ y(t) = 3x(t-1)+x(t+3)-x(t)\! $
Part A
$ h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\! $
$ H(j\omega) = \int_{-\infty}^{\infty}3\delta(t-1)+\delta(t+3)-\delta(t)\,dt\! $
$ H(s) = 3e^{-s} + e^{3s} - 1\! $
Part B
Signal from Question 1
$ x(t)=cos(\frac{\pi}{2}t)u(t) = \frac{1}{2}(e^{j0.5\pi t}+e^{-j0.5\pi t})u(t)\! $
$ y(t) = (3e^{-j0.5\pi} + e^{j1.5\pi} - 1)\times\frac{1}{2}[e^{j0.5\pi t}]u(t)+(3e^{-j0.5\pi} + e^{j1.5\pi} - 1)\times\frac{1}{2}[e^{-j0.5\pi t}]u(t)\! $
$ y(t)=\frac{1}{2}[3e^{j0.5\pi(t-1)}+e^{j0.5\pi(t+3)}-e^{j0.5\pi t}+3e^{-j0.5\pi(t+1)}+e^{j0.5\pi(3-t)}-e^{-j0.5\pi t}]u(t)\! $
